Whakaoti i te f(t)=2t/1+2t^2

$$f ( t ) = \frac { 2 t } { 1 + 2 t ^ { 2 } }$$

Kimi Pārōnaki e ai ki t

$$\frac{2\left(1-2t^{2}\right)}{\left(2t^{2}+1\right)^{2}}$$

Ngā Upane e Whakamahi ana i te Ture Pārōnaki mō te Otinga

Mō ngā pānga e rua e taea ana te pārōnaki, ko te pārōnaki o te otinga o ngā pānga e rua ko te tauraro whakareatia ki te pārōnaki o te taurunga tango i te taurunga whakareatia ki te pārōnaki o te tauraro, ā, ka whakawehea te katoa ki te tauraro kua pūruatia.

$$\frac{\left(2t^{2}+1\right)\frac{\mathrm{d}}{\mathrm{d}t}(2t^{1})-2t^{1}\frac{\mathrm{d}}{\mathrm{d}t}(2t^{2}+1)}{\left(2t^{2}+1\right)^{2}}$$

Ko te pārōnaki o tētahi pūrau ko te tapeke o ngā pārōnaki o ōna kīanga tau. Ko te pārōnaki o tētahi kīanga tau pūmau ko 0. Ko te pārōnaki o te ax^{n} ko te nax^{n-1}.

$$\frac{\left(2t^{2}+1\right)\times 2t^{1-1}-2t^{1}\times 2\times 2t^{2-1}}{\left(2t^{2}+1\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{\left(2t^{2}+1\right)\times 2t^{0}-2t^{1}\times 4t^{1}}{\left(2t^{2}+1\right)^{2}}$$

Whakarohaina mā te āhuatanga tohatoha.

$$\frac{2t^{2}\times 2t^{0}+2t^{0}-2t^{1}\times 4t^{1}}{\left(2t^{2}+1\right)^{2}}$$

Hei whakarea pū o te pūtake ōrite, tāpiri ana taupū.

$$\frac{2\times 2t^{2}+2t^{0}-2\times 4t^{1+1}}{\left(2t^{2}+1\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{4t^{2}+2t^{0}-8t^{2}}{\left(2t^{2}+1\right)^{2}}$$

Pahekotia ngā kīanga tau ōrite.

$$\frac{\left(4-8\right)t^{2}+2t^{0}}{\left(2t^{2}+1\right)^{2}}$$

Tango 8 mai i 4.

$$\frac{-4t^{2}+2t^{0}}{\left(2t^{2}+1\right)^{2}}$$

Tauwehea te 2.

$$\frac{2\left(-2t^{2}+t^{0}\right)}{\left(2t^{2}+1\right)^{2}}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$\frac{2\left(-2t^{2}+1\right)}{\left(2t^{2}+1\right)^{2}}$$

Aromātai

$$\frac{2t}{2t^{2}+1}$$